differentiate sin^2 3x. tan^3 2x
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Answer:
ddxtan2(3x)=6sec2(3x)tan(3x)
Explanation:
In order to differentiate this function, we have to apply the chain rule twice:
ddxtan(f(x))=sec2(f(x))f'(x)
ddx[tan(x)]n=n[tan(x)]n−1sec2x
So, applying these two rules, we get:
ddxtan2(3x)=2tan(3x)sec2(3x)(3)=6sec2(3x)tan(3x)
ddxtan2(3x)=6sec2(3x)tan(3x)
Explanation:
In order to differentiate this function, we have to apply the chain rule twice:
ddxtan(f(x))=sec2(f(x))f'(x)
ddx[tan(x)]n=n[tan(x)]n−1sec2x
So, applying these two rules, we get:
ddxtan2(3x)=2tan(3x)sec2(3x)(3)=6sec2(3x)tan(3x)
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